Unit 8 Electromagnetic design Episode I

USPAS January 2012, Superconducting accelerator magnets

Unit 8Electromagnetic design

Episode I

Helene Felice and Soren PrestemonLawrence Berkeley National Laboratory (LBNL)

Paolo Ferracin and Ezio TodescoEuropean Organization for Nuclear Research (CERN)

USPAS January 2012, Superconducting accelerator magnets Unit 8: Electromagnetic design episode I – 8.2

QUESTIONS

Which coil shapes give a perfect field ?

How to build a sector coil with a real cable, and with good field quality for the beam ? Where to put wedges ? 41° 49’ 55” N – 88 ° 15’ 07” W


CONTENTS

1. How to generate a perfect fieldDipoles: cos, intersecting ellipses, pseudo-solenoidQuadrupoles: cos2, intersecting ellipses

2. How to build a good field with a sector coilDipolesQuadrupoles


1. HOW TO GENERATE A PERFECT FIELD

Perfect dipoles - 1Wall-dipole: a uniform current density in two walls of infinite height produces a pure dipolar field

+ mechanical structure and winding look easy- the coil is infinite- truncation gives reasonable field quality only for rather large

height the aperture radius (very large coil, not effective)

-60

0

60

-40 0 40

+

+-

-

-60

0

60

-40 0 40

+

+-

-

A wall-dipole, cross-section A practical winding with flat cablesA wall-dipole, artist view



Perfect dipoles - 2Intersecting ellipses: a uniform current density in the area of two intersecting ellipses produces a pure dipolar field

- the aperture is not circular- the shape of the coil is not easy to wind with a flat cable

(ends?)- need of internal mechanical support that reduces available

aperture

-60

0

60

-40 0 40

+

+-

-

-60

0

60

-40 0 40

+

+-

-

Intersecting ellipses A practical (?) winding with flat cables



Perfect dipoles - 2Proof that intersecting circles give perfect

fieldwithin a cylinder carrying uniform current j0, the field is perpendicular to the radial direction and proportional to the distance to the centre r:

Combining the effect of the two cylinders

Similar proof for intersecting ellipses

-60

0

60

-40 0 40

+

+-

-

From M. N. Wilson, pg. 28

200 rj

B

sj

rrrj

B y 2coscos

200

221100

0sinsin2 2211

00

rrrj

Bx



Perfect dipoles - 3Cos: a current density proportional to cos in an annulus - it can be approximated by sectors with uniform current density

+ self supporting structure (roman arch)+ the aperture is circular, the coil is compact+ winding is manageable

-60

0

60

-40 0 40

+

+-

-j = j0 cos

An ideal cosA practical winding with one layer and wedges[from M. N. Wilson, pg.

33]

A practical winding with three layers and no

wedges[from M. N. Wilson, pg.

33]

wedgeCable block

Artist view of a cos magnet

[from Schmuser]



Perfect dipoles - 3Cos theta: proof – we have a distribution

The vector potential reads

and substituting one has

using the orthogonality of Fourier series-60

0

60

-40 0 40

+

+-

-j = j0 cos

)cos()( 0 mjj

)cos(2

),(0

00

mm

jA

m

z

1

2

00

00 )](cos[)cos(1

2),(

n

n

z dnmn

jA

1 0

0 )](cos[1

2),(

n

n

z nn

jA



Perfect quadrupolesCos2: a current density proportional to cos2 in an annulus - approximated by sectors with uniform current density and wedges(Two) intersecting ellipses

Perfect sextupoles: cos3 or three intersecting ellipsesPerfect 2n-poles: cosn or n intersecting ellipses

-60

0

60

-40 0 40

++

- - j = j 0 cos 2

- -

+

+

-60

0

60

-40 0 40

+

+

- -

- -

+

+

Quadrupole as an ideal cos2 Quadrupole as two intersecting ellipses


CONTENTS




2. HOW TO BUILD A GOOD FIELD: SECTOR COILS FOR DIPOLES

We recall the equations relative to a current line we expand in a Taylor series

the multipolar expansion is defined as

the main component is

the non-normalized multipoles are

.)(2

)(0

0

zz

IzB

n

ref

refn z

R

R

IC

0

0

2

0

0

0

01

cos

2

1Re

2 z

I

z

IB

-40

0

40

-40 0 40

z 0=x 0+iy 0

x

y

z=x+iy

B=B y+iB x

1

11

00

0

1

1

00

0

22)(

n

n

ref

n

ref

n

n

R

iyx

z

R

z

I

z

z

z

IzB

1

1

1

1

)(n

n

refnn

n

n

refn R

ziAB

R

zCzB



We compute the central field given by a sector dipole with uniform current density j

Taking into account of current signs

This simple computation is full of consequencesB1 current density (obvious)B1 coil width w (less obvious)B1 is independent of the aperture r (much less obvious)

For a cos,

ddjI +

+

-

-

a

w

r

wj

ddj

Bwr

r 2

cos

24 0

2/

0

20

1

a

a

a

sin2cos

22 00

1 wj

ddj

Bwr

r

0

0

0

01

cos

2

1Re

2 z

I

z

IB



A dipolar symmetry is characterized byUp-down symmetry (with same current sign)Left-right symmetry (with opposite sign)

Why this configuration?Opposite sign in left-right is necessary to avoid that the field created by the left part is canceled by the right oneIn this way all multipoles except B2n+1 are canceled

these multipoles are called “allowed multipoles”Remember the power law decay of multipoles with order (Unit 5) And that field quality specifications concern only first 10-15 multipoles

The field quality optimization of a coil lay-out concerns only a few quantities ! Usually b3 , b5 , b7 , and possibly b9 , b11

+

+

-

-

r

w

...)(

4

5

2

31

refref R

zB

R

zBBzB

...101)(

4

4

52

2

34

1refref R

zb

R

zbBzB



Multipoles of a sector coil

for n=2 one has

and for n>2

Main features of these equationsMultipoles n are proportional to sin ( n angle of the sector)

They can be made equal to zero !Proportional to the inverse of sector distance to power n

High order multipoles are not affected by coil parts far from the centre

r

wRjB

ref1log)2sin(

02 a

a

a

a

a

wr

r

nnref

wr

rn

nref

n ddinRj

ddinRj

C 11

01

0)exp(

)exp(

22

n

rwr

n

nRjB

nnnref

n

2

)()sin(2 2210 a



First allowed multipole B3 (sextupole)

for =/3 (i.e. a 60° sector coil) one has B3=0

Second allowed multipole B5 (decapole)

for =/5 (i.e. a 36° sector coil) or for =2/5 (i.e. a 72° sector coil)

one has B5=0

With one sector one cannot set to zero both multipoles … let us try with more sectors !

wrr

jRB

ref 11

3

)3sin(2

0

3

a

33

40

5

11

5

)5sin(

wrr

jRB ref a

+

+

-

-

a

w

r



Coil with two sectors

Note: we have to work with non-normalized multipoles, which can be added together

Equations to set to zero B3 and B5

There is a one-parameter family of solutions, for instance (48°,60°,72°) or (36°,44°,64°) are solutions

wrr

jRB ref 11

3

3sin3sin3sin 123

20

3

aaa

33

123

40

5)(

11

5

5sin5sin5sin

wrr

jRB

ref aaa

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

a1

a2

a3

0)5sin()5sin()5sin(

0)3sin()3sin()3sin(

123

123

aaaaaa



Let us try to better “understand” why (48°,60°,72°) has B3=B5=0

Multipole n sin nClose to 0, all multipoles are positive72° sets B5=0

The wedge is symmetric around 54° preserves B5=0

It sets B3=0 since the slice [48 °,60 °] has the same contribution as [60°,72°]

hoping that now is a bit more clear …

otherwise just blindly trust algebra !

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

B3>0

B3<0

-1.0

-0.5

0.0

0.5

1.0

0 10 20 30 40 50 60 70 80 90

B3

B5

-1.0

-0.5

0.0

0.5

1.0

0 10 20 30 40 50 60 70 80 90

B3

B5

Series9



One can compute numerically the solutions of There is a one-parameter familyInteger solutions

[0°-24°, 36°-60°] – it has the minimal sector width [0°-36°, 44°-64°] – it has the minimal wedge width[0°-48°, 60°-72°][0°-52°, 72°-88°] – very large sector width (not useful)

one solution ~[0°-43.2°, 52.2°-67.3°] sets also B7=0

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

a1

a2

a3

0)5sin()5sin()5sin(

0)3sin()3sin()3sin(

123

123

aaaaaa

0

15

30

45

60

75

90

60 65 70 75 80 85 90a3 (degrees)

Ang

le (

degr

ees)

-3

-2

-1

0

1

2

3

B7

(uni

ts)

a1 a2 Wedge thickness B7a1 a2



We have seen that with one wedge one can set to zero three multipoles (B3, B5 and B7)

What about two wedges ?

One can set to zero five multipoles (B3, B5, B7 , B9

and B11) ~[0°-33.3°, 37.1°- 53.1°, 63.4°- 71.8°]

0)3sin()3sin()3sin()3sin()3sin( 12345 aaaaa





One wedge, b3=b5=b7=0 [0-43.2,52.2-67.3]

Two wedges, b3=b5=b7=b9=b11=0 [0-33.3,37.1-53.1,63.4- 71.8]



Let us see two coil lay-outs of real magnetsThe Tevatron has two blocks on two layers – with two (thin !!) layers one can set to zero B3 and B5

The RHIC dipole has four blocks

0

20

40

60

0 20 40 60x (mm)

y (m

m)

RHIC main dipole - 1995Tevatron main dipole - 1980

0

20

40

60

0 20 40 60x (mm)

y (m

m)



Limits due to the cable geometryFinite thickness one cannot produce sectors of any widthCables cannot be key-stoned beyond a certain angle, some wedges can be used to better follow the arch

One does not always aim at having zero multipoles

There are other contributions (iron, persistent currents …)

RHIC main dipoleOur case with two wedges

0

20

40

60

0 20 40 60x (mm)

y (m

m)



Case of two layers, no wedges

There solutions only up to w/r<0.5There are two branches that join at w/r0.5, in 1 70° and 2 25°Tevatron dipole fits with these solutions

wrwrwrrB

2

11)3sin(

11)3sin( 213 aa

3323315 )2(

1

)(

1)5sin(

)(

11)5sin(

wrwrwrrB aa

0

50

0 50

a1

a2

r

w

0

20

40

60

80

0 20 40 60 80x (mm)

y (m

m)

72o

37o

Tevatron dipole: w/r0.20

0

15

30

45

60

75

90

0.0 0.1 0.2 0.3 0.4 0.5 0.6w/r (adim)

Ang

le (

degr

ees)

a1 - first branch a2 - first branch

a1 - second branch a2 - second branch

a1 a2

a2a1

Tevatron main dipole


CONTENTS




2. HOW TO BUILD A GOOD FIELD: SECTOR COILS FOR QUADRUPOLES

We recall the equations relative to a current line we expand in a Taylor series

the multipolar expansion is defined as

the main quadrupolar component is

the non-normalized multipoles are

.)(2

)(0

0

zz

IzB

1

11

00

0

1

1

00

0

22)(

n

nn

n

n

R

z

z

R

z

I

z

z

z

IzB

1

1

1

1

)()(n

n

nnn

n

n R

ziAB

R

zCzB

n

ref

refn z

R

R

IC

0

0

2

-40

0

40

-40 0 40

z 0=x 0+iy 0

x

y

z=x+iy

B=B x+iB y

20

0

20

02

2cos

2

1Re

2 z

RI

z

RIB refref



We compute the central field given by a sector quadrupole with uniform current density j

Taking into account of current signs

The gradient is a function of w/r For large w, the gradient increases with log w

Field gradient [T/m]Allowed multipoles

ddjI

-50

0

50

-50 0 50

+

+

+

+

--

- -

r w

...)(

9

10

5

62

refrefref R

zB

R

zB

R

zBzB ....101)(

8

8

104

4

64

refref R

zb

R

zbGzzB

refR

BG 2

r

wRjdd

RjB

refwr

r

ref1ln2sin

42cos

28

0

02

0

2 a

a

20

0

20

02

2cos

2

1Re

2 z

RI

z

RIB refref



First allowed multipole B6 (dodecapole)

for =/6 (i.e. a 30° sector coil) one has B6=0

Second allowed multipole B10

for =/10 (i.e. a 18° sector coil) or for =/5 (i.e. a 36° sector coil)

one has B5=0

The conditions look similar to the dipole case …

44

50

6

11

6

)6sin(

wrr

jRB ref a

88

80

10

11

10

)10sin(

wrr

jRB ref a



For a sector coil with one layer, the same results of the dipole case hold with the following transformation

Angels have to be divided by two Multipole orders have to be multiplied by two

ExamplesOne wedge coil: ~[0°-48°, 60°-72°] sets to zero b3 and b5 in dipolesOne wedge coil: ~[0°-24°, 30°-36°] sets to zero b6 and b10 in quadrupoles

The LHC main quadrupole is

based on this lay-out: one wedgebetween 24° and 30°, plus one on the outer layer for keystoning thecoil

0

20

40

0 20 40 60 80x (mm)

y (m

m)

LHC main quadrupole



ExamplesOne wedge coil: ~[0°-12°, 18°-30°] sets to zero b6 and b10 in quadrupoles

The Tevatron main quadrupole is based on this lay-out: a 30° sector coil with one wedge between 12° and 18° (inner layer) to zero b6 and b10 – the outer layer can be at 30° without wedges since it does not

affect much b10

One wedge coil: ~[0°-18°, 22°-32°] sets to zero b6 and b10 The RHIC main quadrupole is based on this lay-out – its is the solution with the smallest angular width of the wedge

Tevatron main quadrupole

0

20

40

0 20 40 60 80x (mm)

y (m

m)

RHIC main quadrupole

0

20

40

0 20 40 60 80x (mm)

y (m

m)


3. CONCLUSIONS

We have shown how to generate a pure multipolar field

We showed cases for the dipole and quadrupole fields

We analysed the constraints for having a perfect field quality in a sector coil

Equations for zeroing multipoles can be givenLay-outs canceling field harmonics can be foundSeveral built dipoles and quadrupoles follow these guidelines

Coming soonPutting together what we saw about design and superconductor limits computation of the short sample field


REFERENCES

Field quality constraintsM. N. Wilson, Ch. 1P. Schmuser, Ch. 4A. Asner, Ch. 9Classes given by A. Devred at USPAS

Unit 8 Electromagnetic design Episode I

Documents

Transcript of Unit 8 Electromagnetic design Episode I